Determinants NCERT Solutions
NCERT Exercise 4.5
Find adjoint of each of the matrices in Exercise 1 and 2.
Question:1. `|(1, 2), (3, 4)|`
Solution:
Let `|(1, 2), (3, 4)|`
`:. A_11=4,A_12=` `-2,A_22+1`
`:.adj A=|(4, -3), (-2, 1)|=` `|(4, -2), (-3, 1)|`
Question:2. `|(1, -1, 2), (2, 3, 5), (-2, 0, 1)|`
Solution:
`|(1, -1, 2), (2, 3, 5), (-2, 0, 1)|`
`A_11=(3-0)=3, A_12`
`=-(2+10)=-12, A_13`
`=(0+6 )=6,A_21`
`=-(-1-0)=1`
`A_22-(1+4)`
`=5,A_23=-(0-2)`
`=2,A_31=(-5-6)`
`=-11,A_32=`
`=-(-5-4)=-1`
`A_33=(3+2)=5`
`:.adj A = [(3, -12, 6), (1, 5, 2), (-11, -1, 5)]`
`=[(3, 1, -11), (-12, 5, -1), (6, 2, 5)]`
Verify A (adj A)A=|A| I in Exercises 3 and 4
Question:3. `[(2, 3), (-4, -6)]`
Solution:
Here A=`[(2, 3), (-4, -6)]`
Now |A|=`|(2, 3), (-4, -6)| `
`=-12-(-12)`
`=-12+12=0`
`A_11=-6,A_12`
`=-(-4)=4, A_21`
`=-3,A_22=2`
`:. adj A= [(-6, 4), (-3, 2)]`
`=[(-6, -3),(4, 2)]`
`A(adj A)=[(2, 3),(-4, -6)]` `[(-6, -3), (4, 2)]`
` =[(-12+12, -6+6), (24-24, 12, 12)]`
`=[(0, 0), (0, 0)]= 0`
`(adj A)A = [(-6, -3), (4, 2)]` `[(2, 3), (-4, -6)]
`=[(-12+12, -18+18),(8-8, 12-12)]`
`=[(0, 0),(0, 0)]=0`
`|A| I =0[(1, 0), (0, 1)]=0`
Thus `A(adj A)=(adj A)A =|A|I`
Question:4 .
`[(1, -1, 2),(2, 3, 5),(-2, 0, 1), (1, 0, 3)]`
Solution:
Let `[(1, -1, 2),(2, 3, 5),(-2, 0, 1), (1, 0, 3)]`
`|A|=[(1, -1, 2),(2, 3, 5),(-2, 0, 1), (1, 0, 3)]`
`=1(0-0)-(-1)` `(9+2)+ 2(0-0)
`=0 +11+ 0=11`
`:. A_11 + (0-0)` `0,A_12=-(9+2)`
`=-11,A_13=(0-0)=0`
`A_21=-(-3-0)`
`3, A_22=(3-2)`
`=1, A_23`
`=-(0+1)=-1`
`A_31(2-0)`
`=2,A_32`
`=-(-2-6)`
`=8,A_33`
`=(0+3)=3`
`adj A=[(0, -11, 0), (3, 1, -1), (2, 8, 3)]`
`=[(0, 3, 2), (-11, 1, 8), (0, -1, 3)]`
`A(adj A)` `= [(1, -1, 2),(3, 0, -2), (1, 0, 3)][(0, 3, 2),(-11, 1, 8),(0, -1, 3)]`
`[(0+11+0, 3-1-2, 2-8+6),(0+0+0, 9+0+2, 6+0-6), (0+0+0, 3+0-3, 2+0+9)]`
`=[(11, 0, 0),(0, 11, 0),(0, 0, 11)]`
`=11[(1, 0, 0), (0, 1, 0),(0, 0, 1)]` `=|A|I_3`
`(adj A)` `=[(0, 3, 2), (-11, 1, 8), (0, -1, 3)]` `[(1, -1, 2), (3, 0, -2), (1, 0, 3)]`
`=[(0+9+2, 0+0s+0, 0+6+6), (-11+3+8, 11+0+0, -22-2+24), (0-3+3, 0+0+0, 0+2+9)]`
`=[(11, 0, 0), (0, 11, 0), (0, 0, 11)]`
`=11[(1, 0, 0), (0, 1, 0), (0, 0, 1)]`
`=|A|I_3`
Thus A` (adj A)A=|A|I_3`
Find the inverse of each of the matrices(if it exists)given in Exercises5 to 11
Question:5. `[(2, -2), (4, 3)]`
Solution:
Let`[(2, -2), (4, 3)]`
|A|=`[(2, -2), (4, 3)]`
`=6-(-8)` `=6+8=14`
`A_11=3,A_12` `=-4,A+21`
`=-(-2)=2,A_22=2`
`adj A= [(3, -4), (2, 2)]`
`=[(3, 2), (-4, 2)]`
`A^-1 = 1/|A| adj A`
`1/(14)[(3, 2), (-4, 2)]`
Question:6 . `[(-1, 5), (-3, 2)]`
Solution:
Let`[(-1, 5), (-3, 2)]`
`|A|[(-1, 5), (-3, 2)]`
`=-2-(-15)`
`=-2+15=13`
`A_11=2,A_12=`
`=-(-3)=3,A_21`
`=-5,A_22=-1`
`adj A=[(2, 3), (-5, -1)]`
`=[(2, -5), (3, -1)]`
`:. A^-1 = 1/|A| adj A`
`=1/13 [(2, -5), (3, -1)]`
Question:7 . `[(1, 2, 3), (0, 2, 4), (0, 0, 5)]`
Solution:
Let`[(1, 2, 3), (0, 2, 4), (0, 0, 5)]`
`|A|[(1, 2, 3), (0, 2, 4), (0, 0, 5)]`
`=1(10-0)-2(0-0)` `+3(0-0)`
`10-0+0=10`
`A_11=(10-0)=10,A_12`
`=-(0-0)=0,A_13` `=(0-0)=0`
`A_21=(10-0)=-10,A_22`
`=(5-0)=5,A_23` `=-(0-0)=-(0-0)=0`
`A_31=(8-6)=2,A_32`
`=-(4-0)=-4,A_33` `=(2-0)=2`
`adj A=[(10, 0, 0),(-10, 5, 0), (2, -4, 2)]`
`=[(10, -10, 2),(0, 5, -4), (0, 0, 2)]`
`A^-1 =1/|A| adj A`
`1/10 [(10, -10, 2),(0, 5, -4), (0, 0, 2)]`
Question:8 . `[(1, 0, 0), (3, 3, 0), (5, 2, -1)]`
Solution:
Let A =`[(1, 0, 0), (3, 3, 0), (5, 2, -1)]`
`|A|=[(1, 0, 0), (3, 3, 0), (5, 2, -1)]`
`1(-3-0)-0(-3-0)` `+0(6-15)`
`=-3-0+0=-3`
`A_11=(-3-0)`
`=-3,A_12`
`=-(-3-0)``= 3, A_13`
`=(6-15)=-9`
`A_21=-(0-0)0,A_22`
`=(1-0)=-1,A_23`
`=-(2-0)=-2`
`A_31=(0-0)=0,A_32`
`=-(0-0)=0, A_33`
`=(3-0)=3`
`adj A[(-3, 3, -9), (0, -1, -2), (0, 0, 3)]`
`= [(-3, 0, 0), (3, -1, 0), (-9, -2, 3)]`
`A^-1=1/|A|adj A`
`=1/-3[(-3, 0, 0), (-3, 1, 0), (9, 2, -3)]`
`1/3= [(3, 0, 0), (-3, 1, 0), (9, 2, -3)]`
Reference: